Bob and Wendy left home to walk together to a restaurant for

Question

Bob and Wendy left home to walk together to a restaurant for dinner. They started out walking at a constant pace of 3 mph. At precisely the halfway point, Bob realized he had forgotten to lock the front door of their home. Wendy continued on to the restaurant at the same constant pace. Meanwhile, Bob, traveling at a new constant speed on the same route, returned home to lock the door and then went to the restaurant to join Wendy. How long did Wendy have to wait for Bob at the restaurant?

(1) Bob’s average speed for the entire journey was 4 mph.
(2) On his journey, Bob spent 32 more minutes alone than he did walking with Wendy.

Guys - any idea how to solve this? Also, can someone please explain how should I approach to solve these questions?

(1) Bob’s average speed for the entire journey was 4 mph. 
(2) On his journey, Bob spent 32 more minutes alone than he did walking with Wendy.

Manhnip · Accepted Answer

This question cannot necessarily be rephrased, but it is important to recognize that we need not necessarily calculate Wendy’s or Bob’s travel time individually. Determining the difference between Wendy’s and Bob’s total travel times would be sufficient. This difference might be expressed as tb – tw.

(1) INSUFFICIENT: Calculating Bob’s rate of speed for any leg of the trip will not give us sufficient information to determine the time or distance of his journey, at least one of which would be necessary to determine how quickly Wendy reaches the restaurant.

(2) SUFFICIENT: To see why this statement is sufficient, it is helpful to think of Bob's journey in two legs: the first leg walking together with Wendy (t1), and the second walking alone (t2). Bob's total travel time tb = t1 + t2. Because Wendy traveled halfway to the restaurant with Bob, her total travel time tw = 2t1. Substituting these expressions for tb – tw,
t1 + t2 – 2t1 = t2 – t1 tb –tw =t2 –t1

Statement (2) tells us that Bob spent 32 more minutes traveling alone than with Wendy. In other words, t2 – t1 = 32. Wendy waited at the restaurant for 32 minutes for Bob to arrive.

The correct answer is B.

mau5 · Answer

Just some minor glitch there, not affecting the answer in any way though.

Let the total distance be 6d miles

From F.S 1, we know that time taken by Wendy  = \frac{6d}{3} = 2d hours.

Again, we know that Total time taken by Bob  = \frac{Total Distance}{Average Speed} = \frac{(3d+3d+6d)}{4} = \frac {12d}{4} = 3d hours

Thus, the time for which Wendy waited = 3d-2d = d hours. Clearly, this depends on the initial distance.Insufficient.

From F.S 2, we know that "On his journey, Bob spent 32 more minutes alone than he did walking with Wendy "

Time spent by Bob with Wendy =  \frac{3d}{3} = d hours

Time spent by Bob alone =  d+\frac{32}{60} = d+\frac{8}{15} hours

Total time taken by Bob = 2d+\frac{8}{15} hours

Total time taken by Wendy = \frac{6d}{3}= 2d  hours
Thus, time Wendy waited for Bob  = 2d-2d+\frac{8}{15} = \frac{8}{15}  hours

Sufficient.

alpha_plus_gamma · Answer

distance/rate = time Time taken by Wendy t1 = d/3 Total Time taken by Bob t2 = d/(2*3) + d/2x + d/x = d (1/6 + 1/2x + 1/x) (1) 1/6 + 1/2x + 1/x = 4 x/6x + 3/6x + 6/6x = 4 x + 9= 24x ; x = 9/23, this is bob's new speed with this the time Wendy waited can be calculated. (2) this does not help to know the time for half-return-distance by Bob. So, its (1) alone is sufficient. (I am not at all sure though, :lol:

)

walker · Answer

In 1 we can double all distances and time also will be doubled because the average speed is determined by ratio of distances rather than their absolute values.

You can simply imagine the situations when distance between home and restaurant is 10 meters and when it is 1000 km

walker · Answer

Bob and Wendy left home to walk together to a restaurant for dinner. They started out walking at a constant pace of 3 mph. At precisely the halfway point, Bob realized he had forgotten to lock the front door of their home. Wendy continued on to the restaurant at the same constant pace. Meanwhile, Bob, traveling at a new constant speed on the same route, returned home to lock the door and then went to the restaurant to join Wendy. How long did Wendy have to wait for Bob at the restaurant?

(1) Bob’s average speed for the entire journey was 4 mph.
(2) On his journey, Bob spent 32 more minutes alone than he did walking with Wendy.

Guys - any idea how to solve this? Also, can someone please explain how should I approach to solve these questions?

(2) On his journey, Bob spent 32 more minutes alone than  he did walking with Wendy. 
---> than  Wendy spent walking from home to halfway point .--->
(because the same distance and the same constant pace of Wendy) ---> than  Wendy spent walking from halfway point to restaurant .-------->32min

piyatiwari · Answer

Let

Bob's initial speed B = Wendy's initial speed W = 3 MPH

Total distance from home to restaurant = D

Bobs new speed NB

So,

1. Total time Wendy took to reach restaurant = D/3 hours.

2. Bob's travel = (D/2 miles With wendy at 3 MPH) + (1.5 D at NB)

So total time bob took to reach restaurant = (D/2)/3 + 1.5D/NB = D/6 + 1.5D/NB = D (1/6 + 1.5/NB)

What is asked : How long did Wendy have to wait for Bob at the restaurant?

= Time Bob took - Time wendy took.
= D (1/6 + 1.5/NB) - D/3

So if we know D and NB we will be able to find answer.

Given statements:

(1) Avg speed = 4 MPH

=> total distance / total speed = 2D / (3+NB) = 4

(2) (D/2)/3 + 32 = 1.5D/NB

Hence both statements together are necessary.

Bunuel · Answer

Check similar problem here: bob-and-wendy-planned-to-walk-from-their-home-to-a-89738.html#p679745 Hope it helps.

budablasta · Answer

Hold on. (3+NB) doesn't seem ok. You can't  add  speeds. You can add distances and times, but not speeds.

Bunuel · Answer

You can add or subtract rates to get relative rate.

For example if two cars are moving toward each other from A to B (AB=100 miles) with 10mph and 15mph respectively, then their relative (combined) rate is 10+15=25mph, and they'll meet in time=distance/rate=100/25=4 hours;

Similarly if car x is 100 miles ahead of car y and they are moving in the same direction with 10mph and 15mph respectively then their relative rate is 15-10=5mph, and y will catch up x in 100/5=20 hours.

Hope it's clear.

WholeLottaLove · Answer

Bob and Wendy left home to walk together to a restaurant for dinner. They started out walking at a constant pace of 3 mph. At precisely the halfway point, Bob realized he had forgotten to lock the front door of their home. Wendy continued on to the restaurant at the same constant pace. Meanwhile, Bob, traveling at a new constant speed on the same route, returned home to lock the door and then went to the restaurant to join Wendy. How long did Wendy have to wait for Bob at the restaurant?

(1) Bob’s average speed for the entire journey was 4 mph.
(2) On his journey, Bob spent 32 more minutes alone than he did walking with Wendy.

Guys - any idea how to solve this? Also, can someone please explain how should I approach to solve these questions?

(1) Bob’s average speed for the entire journey was 4 mph. 
While we know that his average speed was 4 miles/hour, the answer to the question depends on the distance from home to diner. Let's say that the distance was 8. Furthermore we can break it up into two, 4 mile segments.

Time = Distance/Rate
Wendy's Time = 8/3 hours or roughly 160 minutes.

Now Bob traveling at his faster rate would cover 4 miles in an hour. However, Bob would need to travel the 4 miles with Wendy, 4 miles back home, then 8 miles to the diner meaning he covered 16 miles in total. He would have spent 4 hours (240 minutes) walking so she would have spent 240-160 = 80 minutes waiting.

Now lets pretend the diner was 4 miles away.

Time = Distance/Rate
Wendy Time = 4/3 hours or roughly 75 minutes.

Bob would travel 2 miles (the half way point), two miles back home then another 4 miles from home to the diner for a total of 8 miles. At 4 MPH average this would take 2 hours (120 minutes) meaning she waited 120-75 = 35 minutes.
INSUFFICIENT

(2) On his journey, Bob spent 32 more minutes alone than he did walking with Wendy.

Bob and Wendy left at a constant rate. Let's say the distance from home to the Diner was 4 miles. This means that T=d/r and Wendy took 4/3rds hours (75 minutes) to get from home to the diner. Bob would have traveled 37.5 minutes with Wendy before he turned around to go lock the door. If he spent 37.5 minutes with Wendy then he would have spent 37.5+32 = 69.5 minutes traveling from midpoint to home to the restaurant. During this time Wendy would have walked an additional 37.5 minutes to get to the diner. The time she spent alone would have been 69.5-37.5 = 32 minutes alone.

Just to prove distance doesn't matter...

Bob and Wendy left at a constant rate. Let's say the distance from home to the Diner was  40 miles . This means that T=d/r and Wendy took 40/3rds hours (750 minutes) to get from home to the diner. Bob would have traveled 375 minutes with Wendy before he turned around to go lock the door. If he spent 375 minutes with Wendy then he would have spent 375+32 = 407 minutes traveling from midpoint to home to the restaurant. During this time Wendy would have walked an additional 375 minutes to get to the diner. This time she spent alone would have been 407-375 = 32 minutes alone.

The additional 32 minutes he spent alone represents his rate traveling from midpoint to home to midpoint to diner. If the distance is greater than his rate will increase but the time it takes him to make the trip will be the same (because the increase in distance is matched by the increase in rate.)

The trick here is to realize that he spent 32 more minutes alone than he did with her. Bob went half way with Wendy which means he spent that amount of time plus 32 minutes going back home and to the diner. 
SUFFICIENT

(B)

I would LOVE to see the algebraic explanation if possible!

Thanks!

semwal · Answer

my soln;

let distance = x
time taken to travel half the distance = x/6 hours. wendy reaches end in x/3 and waits.
let bob,s new speed k. total time travelled by bob upto reaching end of journey- x/6+ 3x/2k 
hence wendy waited x/6 +3x/2k - x/3 =  3x/2k - x/6 hours.

statement 1. gives k but not x hence insufficient.
statement 2.  3x/2k - x/6= 32 minutes  exactly what we want . hence sufficient.

Ans = B

jlgdr · Answer

Good question +1 Let's say that total distance is 24 miles Statement 1 Now, Bob's average speed for entire journey was 4mph. Since we have different rates at which he traveled namely, 3mph and 'x' we cannot find the total time he took unless we have that 'x' rate Statement 2 Here we get that we spend 32 more minutes walking alone that with Wendy So if distance is 24 he spend 12/3 = 4 hours walking with wendy and then he spent (12+24=36/x) hours walking alone where 'x' is the second speed at which he travelled We have a linear equation with one variable hence, we can solve for 'x' This is sufficient Answer is thus B Does it all make sense? Cheers! J

aditya27cs08 · Answer

I see it , this one reminded me of my school days .

First of all analyze the equation that can be formed from the question itself . What do we need to find out ? If T1 = The time taken by wendy to reach the restaurant and T2 = The time taken by Bob to do the same Then we need to find T2 - T1 ( bob took extra time due to forgetfulness) Now , we form equations Let Distance = x Then , T1 = x/3 Hours T2 = 1/2 of x Divided by 3 PLUS 1/2 of x Divided by Y(new speed returning halfway) PLUS x Divided by Y ( from home to restaurant on new speed) When we do T2-T1 we get , (9x-xy)/6y Now statement 2 says that : BOB walked 32 minutes more alone than he walked with wendy . Bob walked alone only when he went halfway back to fix the lock and then full way to the restaurant = 1/2 of x Divided by Y PLUS x/Y Bob walked with wendy only for halfway of x = 1/2 of x Divided by 3(then speed) Difference is = 32/60 hours therefore , (1/2 of x Divided by Y PLUS x/Y) - (1/2 of x Divided by 3) = 32/60 Solving this we will get the value of xy = 9x - 3.2y Put it in previous equation (9x-xy)/6y . We get 8/15 hours . Hence Statement 2 is enough . Hope it helps !!

syntnghosh · Answer

A)   let total distance be 2D (D for halfway)
Time taken by Wendy = 2D/3
Time taken by Bob = D/3 + 3D/Vb

average speed of Bob = 4D/(D/3 + 3D/Vb)= 12Vb/(Vb+9) = 4; solving we get Vb= 9/2
but we don't know the distance so the time cannot be calculated.
 Insufficient

B)   Time taken by Wendy = (t1 + t1)
Time taken by Bob = t1 + 3t2
Difference in time, which is the waiting time = (3t2 +t1) - (t1 + t1)= 3t2 - t1
Also given  Bob spent 32 more minutes alone than he did walking with Wendy 
Time Bob spent with Wendy, t1
Time Bob spent alone = 3t2
Difference = 3t2 - t1 = 32 mins
since 3t2 - t1 = 32 is the waiting time as shown above.
 Sufficient

szhuge · Answer

Answer is B.  Quick way to reason this without doing any calculations:

1)  Insufficient.  Knowing the average speed of the entire journey will let you calculate Bob's new speed, but it will not let you calculate any absolute time or distance estimates (which is what we need to calculate the difference between Bob's total time and Wendy's total time.)

This is because  the average speed of two different speeds depends only on the proportion of the total time or distance dedicated to each speed, but not on any absolute estimates of time or distance.  For instance, if I doubled the distance, Bob's average speed would stay the same, but Wendy would have to wait longer; therefore average speed is not sufficient information.

2)  Sufficient. In fact, the answer to the overall question is 32 minutes.  The question essentially asks for the difference between Bob's total time and Wendy's total time. Since we're only looking for the difference in total times, we can "subtract out" Bob/Wendy's joint time for both people;  thus the question essentially asks for the difference between Bob's alone time and Wendy's alone time.

Since Wendy was walking at a constant speed, and since Bob left Wendy halfway, we know that Wendy spent half her time with Bob and half her time alone. Therefore  Bob/Wendy's joint time is equal to Wendy's alone time.

Statement (2) tells us that Bob's alone time was 32 minutes longer than Bob/Wendy's joint time, therefore Bob's alone time was 32 minutes longer than Wendy's alone time, answering the overall question.

Himanshu_Taneja · Answer

B can't be sufficient as B doesn't consider the time wendy took to reach restraunt, question mentions how much time wendy waited at the restraunt. so that would be 32 mins - time wendy would take to reach the restraint

as given 32 mins is time wendy is alone i.e time wendy took from mid point to restraunt(equal to time wendy took from home to mid point) + time wendy waited at the restraunt

Fdambro294 · Answer

Statement 1: 
Let us call the distance from home to restaurant, measures in miles = D

We are looking for the Time that Wendy spent waiting for Bob.

This would be:
(Bob travel time). - (Wendy travel time)

Time = distance / rate

Bob travels twice the distance from the home to the restaurant (he travels halfway, then back, and then another full distance).

His average speed = 4 mph

Bob time = 2D / 4 = D/2

Wendy travels the entire distance D at 3 mph

Wendy time = D/3

(D/2) — (D/3) = D/6 hours that Wendy must wait

We would need the distance (other information would suffice as well)

S2:

The trap is that most students will start to try to algebraically manipulqte the data and not step back to think logically about the given information and the target question.

We need to know how much time Wendy waited.

Another way to state this, we need to know the extra time that Bob spent walking after Wendy arrived at the restaurant.

Stated another way, we need to know the time that Bob spent walking ALONE, because Wendy arrived at the restaurant.

The statement is tricky, but Bob and Wendy walk TOGETHER up until the point Wendy arrives at the restaurant, even if Bob is no longer traveling in the same direction, side by side with Wendy.

So, Wendy will have to wait 32 minutes (Wendy waits the time that Bob spends waking alone in either direction)

*B*

Note: I answered this incorrectly the first time as well. Tricky wording and tricky question.

Posted from my mobile device

agarwal6820 · Answer

Hello,

Thank you for the solution, however are you not overlooking the fact that Wendy will also take t1 time to reach the restaurant from halfway, so let's spread out the situation, BOB spends 32 minutes alone without Wendy, which means that when he leaves halfway till the time he reaches the restaurant, the time spanned is 32 mins. In this time Wendy also has to cover the half distance in which she will get to the restaurant and then wait at the restaurant. So, t1(time for Wendy to cover half distance to the restaurant) + time Wendy waits at the restaurant should be equal to 32 minutes, SOOOOOO, time Wendy waits at the restaurant will definitely be less than 32 minutes. It will lie between 0 to 32 minutes. If Bob travels at 3 times the speed when going back they both will reach together, hence wait time would be 0. IF Bob travels any slower the wait time will increase but cannot exceed 32 minutes considering the 2nd option given in the question. :: 0<= wait time of Wendy at restaurant< 32.

If you take 32 minutes as Wendy's wait time, that means Wendy covered the second half in 0 seconds making her speed infinite, which is inconsistent.

Hope I am making sense.

Oppenheimer1945 · Answer

Let total dist be 2d

What is (3d/v)-d/3=3d/v-d/3=?

S1) 4d/(d/3+3d/v)=4=
=>v=... (NS)

S2) 3d/v-d/3=32min
(Suff)
B)

Bob and Wendy left home to walk together to a restaurant for

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